If I have a curve given by $$ y^2 = (x^3-1)(x^3-a), $$ how do I find out if there is a rational variable transformation $y=y(s,t)$, $x=x(s,t)$ that maps this curve onto an elliptic curve of the form $$ s^2+a_1st+a_3 s=t^3+a_2t^2+a_4t+a_6. $$ I think it's relevant that $x\mapsto a^{1/3}/x$ leaves the roots of the r.h.s. unchanged, but I don't know how I can use that. I learned that there should exist a quotient of the original curve by the involution $x\mapsto a^{1/3}/x$, but I am not clear how I would actually compute it.

Is there a way to explicitly determine the rational functions $y(s,t)$, $x(s,t)$?


I think that taking the quotient amounts to the following. I think about this at the level of function fields. So let $k$ be the field of constants. Then the function field of your curve is $F=k(x,y)$, which is a quadratic extension of the field of rational functions $K=k(x)$, where $y^2=(x^3-1)(x^3-a)$.

The symmetry that you talk about can be viewed as an automorphism $\phi$ of the field $F$. To avoid fractional exponents I denote $a=A^6$. Here I need to assume that $a$ has a sixth root in $k$. Hopefully this will not pose a problem. Now we can define the $k$-automorphism of $F$ by the rules $$ \phi:x\mapsto\frac{A^2}x,\qquad y\mapsto \frac{yA^3}{x^3}. $$ We need to verify that this is an automorphism of $F$. This is equivalent to checking that $\phi(y)^2=(\phi(x)^3-1)(\phi(x)^3-A^6)$, and verifying that is straightforward.

You hinted at the possibility that we are moding out an involution. This is the case, because we easily see that $$ \phi(\phi(x))=x,\qquad\text{and}\qquad\phi(\phi(y))=y. $$ Thus $\phi$ generates a cyclic group of order two. Basic Galois theory then tells us that $L=\operatorname{Inv}(\phi)$ is a subfield of $F$ such that $[F:L]=2$. Our next task is to identify the fixed field $L$. The first and most obvious observation is that $$ u=x+\phi(x)=x+\frac{A^2}x\in L. $$ This is immediate from the involutory property of $\phi$. At this point we observe that $x$ is algebraic of degree two over the field $M=k(u)$. Thus we conclude that $M=L\cap K$.

Finding another "useful" element in $L$ was a bit tricky. It is easy enough to find elements from $L\setminus M$, but they were kludgy to work with. A choice that worked for me is $$ v=\frac yx+\phi(\frac yx)=y\left(\frac1x+\frac A{x^2}\right). $$ Clearly $v^2$ will be an element of $M$ - we just need to calculate it $$ \begin{aligned} v^2&=\frac{y^2}{x^2}\left(1+\frac Ax\right)^2=\frac{(x^3-1)(x^3-A^6)}{x^2}\left(1+\frac Ax\right)^2\\ &=\frac{A^8 + 2 A^7 x + A^6 x^2 - A^2 x^3 - A^8 x^3 - 2 A x^4 - 2 A^7 x^4 - x^5 - A^6 x^5 + A^2 x^6 + 2 A x^7 + x^8}{x^4}\\ &=u^4 + 2 A u^3 - 3 A^2 u^2 - (A^6 + 6 A^3 + 1) u-2(A^7+A).\\ \end{aligned} $$

It is easy to see that $L=M(v)=k(u,v)$. Evidently $k(u,v)\subseteq L$. Also clearly $y\in k(u,v,x)$, so $F=k(u,v,x)$ and we saw that $x$ is quadratic over $k(u,v)$ so $[F:k(u,v)]=2$. As $[F:L]=2$ the claim follows.

From the general theory of hyperelliptic curves we infer that the equation $$ v^2=u^4 + 2 A u^3 - 3 A^2 u^2 - (A^6 + 6 A^3 + 1) u-2(A^7+A)\qquad(*) $$ defines a curve of genus $g=1$ whenever the roots of that quartic are distinct. Therefore $L$ is the function field of an elliptic curve whenever that holds.

The process of transforming the curve $(*)$ to a cubic needs to move one of the zeros of the r.h.s. to the infinity. Here we observe that $u=-2A$ is a zero of that quartic, so let's work with that. We can factor $$ u^4 + 2 A u^3 - 3 A^2 u^2 - (A^6 + 6 A^3 + 1) u-2(A^7+A)=(u+2A)(u^3-3A^2u-A^6-1). $$ Here by Taylor expansion around $u=-2A$ $$ u^3-3A^2u-A^6-1=(u+2A)^3-6A(u+2A)^2+9A^2(u+2A)-(A^3+1)^2. $$ This implies that dividing $(*)$ by $(u+2A)^4$ gives us, in the variables $$ w=\frac{v}{(u+2A)^2},\qquad z=\frac1{u+2A}, $$ the equation $$ w^2=1-6Az+9A^2z^2-(A^3+1)^2z^3.\qquad(**) $$ Getting warmer! To get to the Weierstrass form we multiply $(**)$ by $(A^3+1)^4$ and introduce, finally, the variables $$ s=(A^3+1)^2w,\qquad t=-(A^3+1)^2z, $$ whence the equation takes the form $$ s^2=t^3+9A^2t^2+6A(A^3+1)^2z+(A^3+1)^4. $$

| cite | improve this answer | |
  • $\begingroup$ Thank you very much. I was trying to find an integral of the form $\int dx/y$, and I read somewhere that I could relate it to $\int du/v$ by taking the quotient by the involution. It was a frustrating problem because I though it should be simple, but all the books I could find omitted the direct calculations entirely and I didn't know enough background to reconstruct them myself. $\endgroup$ – Kirill Oct 12 '14 at 19:26
  • $\begingroup$ Looks good except for $-(A^3+1)z^3$: I get $-(A^3+1)^2z^3$ instead. $\endgroup$ – Kirill Oct 12 '14 at 19:34
  • $\begingroup$ Better now? ${}$ $\endgroup$ – Jyrki Lahtonen Oct 12 '14 at 19:38
  • $\begingroup$ Yes. Thank you so much, this is a lot clearer to me now. $\endgroup$ – Kirill Oct 12 '14 at 19:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.